Slog(Exponential Factorial(x))

[tommy1729]

EF(x) = exponential factorial(x) = x^(x-1)^(x-2)^...^3^2^1.
What happens if you do the tetration logarithm of the exponential factorial function. (I am thinking tetration logarithm base the Tetra-Euler Number.) How can the tetration logarithm of the exponential factorial function be approximated?
slog(e4,EF(1)) = 0.
slog(e4,EF(2)) ~ .636.
slog(e4,EF(3)) ~ 1.612.
slog(e4,EF(4)) ~ 2.693.
slog(e4,EF(5)) ~ 3.703.
Numbers worked out with the Kneser method.
This is my 16th thread! 
I conjecture that slog(k,EF(x))-x approaches a number, as x grows larger and larger. For any real k greater than eta.

Let FE(b,x) = b^3^4^...^x.
FE(e,x) = FE(x) 

Lets get some rude boundaries clear.

 x^e^e^... < x^(x-1)^(x-2)^... < e^e^...^x < e^3^4^...^x <  x^x^...^x

taking slog on both sides.

1 + slog( e^e^... * ln(x) ) < slog (x^(x-1)^(x-2)^...) < x + slog(x) < slog(FE(x)) <  x + basechange(e,x)

2 + slog( e^e^... * ln(ln(x)) ) < slog (x^(x-1)^(x-2)^...) < x + slog(x) < slog(FE(x)) < x + basechange(e,x)

...

x + slog(ln ln ln ... ln x  ) < ...

upperbound for x + slog( ln ln ln ... ln x) = x + 0.99

So we end up with the estimated

x + 0.99 < slog( EF(x) ) < x + slog(x).

trivial to say but slog(EF(x)) / x thus clearly converges.

It is not immediately clear how to continue.

How does one show that slog( EF(x) ) < x  + slog(slog(slog(x))).

or that  ln^[1/2] ( slog( EF(x) ) ) - ln^[1/2](x) converges.   

Maybe the answer lies in one of the thousands posts here.

Or someone else might find it.


regards

tommy1729